W WEMG Member Joined Aug 15, 2009 Messages 118 Gender Undisclosed HSC 2011 Dec 17, 2010 #1 how do i simplify e^lnx so that I can differentiate it?
I iSplicer Well-Known Member Joined Jun 11, 2008 Messages 1,809 Location Strathfield Gender Male HSC 2010 Uni Grad 2017 Dec 17, 2010 #2 WEMG said: how do i simplify e^lnx so that I can differentiate it? Click to expand... e^ln(x) = x raising e to some value is the exact opposite of taking the base e logarithm of that same value. So when you do e^ln(y), you're cancelling the operations out effectively, leaving x so d(e^ln[x])/dx = 1
WEMG said: how do i simplify e^lnx so that I can differentiate it? Click to expand... e^ln(x) = x raising e to some value is the exact opposite of taking the base e logarithm of that same value. So when you do e^ln(y), you're cancelling the operations out effectively, leaving x so d(e^ln[x])/dx = 1
W WEMG Member Joined Aug 15, 2009 Messages 118 Gender Undisclosed HSC 2011 Dec 17, 2010 #3 Yep, I just realised it after doing another question haha
Pwnage101 Moderator Joined May 3, 2008 Messages 1,408 Location in Pursuit of Happiness. Gender Undisclosed HSC N/A Dec 19, 2010 #4 Above is correct, but just remember: For x>0, exp{ln(x)}=x. For x<=0, exp{ln(x)} is not defined. [Whereas ln(exp{x})=x for all real x.] *Here exp{.}=e^.
Above is correct, but just remember: For x>0, exp{ln(x)}=x. For x<=0, exp{ln(x)} is not defined. [Whereas ln(exp{x})=x for all real x.] *Here exp{.}=e^.
Gussy Booo Mathematics <3 Joined Aug 1, 2009 Messages 251 Location Sydney Gender Male HSC 2010 Dec 19, 2010 #5 May I just add that this result is produced due to the fact that the Exponential and Logarithmic function are INVERSE FUNCTIONS.
May I just add that this result is produced due to the fact that the Exponential and Logarithmic function are INVERSE FUNCTIONS.